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InterviewSolution
This section includes InterviewSolutions, each offering curated multiple-choice questions to sharpen your knowledge and support exam preparation. Choose a topic below to get started.
| 101. |
According to classical theory, the circular path of an electron in Rutherford atom isA. SpiralB. CircularC. ParabolicD. Straight line |
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Answer» Correct Answer - A |
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| 102. |
Assertion: According to classical theory, the proposed path of an electron in Rutherford atom model will be parabolic. Reason: According to electromagnetic theory an accelerated particel continuosly emits radiation.A. If both assertion and reason are true and the reason is the correct explanation of the assertionB. If both assertion and reason are true but reason is not the correct explanation of the assertionC. If assertion is true but reason is false.D. If assertion is false but reason is true |
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Answer» Correct Answer - D |
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| 103. |
The energy of electron in first excited state of `H`-atom is `-3.4 eV` its kinetic energy isA. `-3.4 eV`B. `+ 3.4 eV`C. `-6.8 eV`D. `6.8 eV` |
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Answer» Correct Answer - B |
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| 104. |
Calculate the angular frequency of an electron occuppying the second Bohr orbit of `He^(+)` ion. |
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Answer» This has been calculated before in problem (6.20). It is `omega=(m(Ze^(2)//4pi epsilon_(0))^(2))/( ħ^(3)n^(3))=2.08xx10^(16)rad//sec` |
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| 105. |
Finf the wavelength of the first line of the `HE^(+)`ion spectral series whose interval between the extreme line is `Delta omega= 5.18.10^(15) s^(-1)`. |
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Answer» We start from the generalized Balmer formula `omega=RZ^(2)((1)/n^(2)-(1)/(m^(2)))` Here `m=n+1, n+2,....oo` Then the angular frequency of the first line of this series (series `n`) is `omega_(1)=RZ^(2)((1)/(n^(2))-(1)/((n+1)^(2)))=Delta omega(((n+1)/(n))^(2)-1)` `=Delta omega [{(Zsqrt((R)/(Delta omega)))/(Zsqrt((R)/(Deltaomega)))}-1]` Then the wavelength will be `lambda_(1)=(2pi c)/(omega_(1))=(2pi c)/(Delta omega)((Zsqrt((R)/(Delta omega)-1)))/(2Zsqrt((R)/(Delta omega))-1)` Substitution (with the value `R` from problem 6.34 which is also the correct value determined directly) gives `lambda_(1)= 0.468mu m`. |
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| 106. |
For the case of atomic hydrogen find: (a)the wavelength of the first three lines of the Balmer series, (b)the minimum resolving the first 20 lines of the Balmer series. |
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Answer» The energies are `E_(H)((1)/(4)-(1)/(9))=(5)/(36)E_(H),E_(H)((1)/(4)-(1)/(16))=(3)/(16)E_(H)` `E_(H)((1)/(4)-(1)/(25))=(21)/(100)E_(H)` They correspond to wavelengths `654.2nm, 484.6nm` and `433nm` The `n^(th)` line the Balmer series has the energy `E_(H)((1)/(4)-(1)/((n+2)^(2)))` For `n= 19`, we get the wavelength `366.7450nm` For `n= 20` we get the wavelength `366.4470nm` To resolve these lines we require a resolving power of `R~~(lambda)/(delta lambda)=(366.6)/(0.298)= 1.23xx10^(3)` |
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| 107. |
Find the minimum magnitude of the magnetic field induction `B` at which a spectral instrument with resolving power `lambda//delta lambda==1.0.10^(5)` is capable of resolving the components of the spectral line `lambda= 536 nm` caused by a transition between singlet terms. The observation line is at right angles to the magnetic field direction. |
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Answer» From the previous problem, if the components are `lambda,lambda+-Delta lambda`, then `(lambda)/(Delta lambda)=(2pi ħc)/(mu_(B)B lambda)` For resolution `(lambda)/(Delta lambda)leR=(lambda)/(delta lambda)` of the instrument. Thus `(2 pi ħc)/(mu_(B)B lambda)leR or Bge(2pi ħc)/(mu_(B)lambdaR)` Hence the minimum megnetic induction is `B_(m in)=(2pi ħc)/(mu_(B)lambdaR)= 4kG= 0.4T` |
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| 108. |
Calculate the magnetic field induction at the centre of a hydrogen atom caused by an electron moving along the first Bohrorbit. |
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Answer» The revolving is electron is equivalent to a circular current `I=(e )/(T)=(e )/(2pi r//v)=(ev)/(2pir)` The magnetic induction `B=(mu_(0)I)/(2r)=(mu_(0)ev)/(4pir^(2))=(mu_(0))/(4pi).e.(e^(2))/((4pi epsilon_(0)) ħ).[(me^(2))/( ħ^(2)(4pi epsilon_(0)))]^(2)` `=(mu_(0)m^(2)e^(7))/(256pi^(4)epsilon_(0)^(3) ħ^(5))` Substitution gives `B= 12.56T` ar the centre. (In Gaussian units `B=(m^(2)e^(7))/(c ħ^(5))= 125.6kG.`) |
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| 109. |
The effective cross section of a uranium nucleus corresponding to the scattering of monoenergetic alpha particles with an the angular interval from `90^(@)` to `180^(@)` is equal to `Delta sigma= 0.50kb`. Find , (a) the energy of alpha particles, (b) the differntial cross section of scattering `d singma//dOmega(kb//sr)` coresponding to the angle `theta=60^(@)` |
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Answer» (a) From the previous formula `Delta sigma=((Ze^(2))/((4piepsilon_(0))2T))^(2) pi "cot"^(2)(theta_(0))/(2)` or `T=(Ze^(2))/(4pi epsilon_(0))"cot" (theta_(0))/(2)sqrt((pi)/(Deltasigma))` Substituting the values with `Z= 79` we get `(theta_(0)= 90^(@))` `T= 0.903MeV` The differential scattering cross section is `(d singma)/(d Omega)= C "cosec"^(4)(theta)/(2)` where `Delta sigma(theta gt theta_(0))= 4pi C "cot"^(2)(theta_(0))/(2)` Thus from the given data `C=(500)/(4pi)b= 39.79b//sr` So `(dsigma)/(d Omega)(theta=60^(@))= 39.79xx16b//sr= 0.637kb//sr` |
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| 110. |
In accordance with classical electrodynamic an electron moving with acceleration `W` loses its energy due to radiation as `(dE)/(dt)= -(2e^(2))/(3c^(3))W^(2)`, Where `e` is the electron, `c` is the velocity of light. Estimate the time during which the energy of an electron performing alomist harmonic oscillations with frequency `omega= 5.10^(15)s^(-1)` will decrease `eta= 10 time s` |
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Answer» The formula in `MKS` units is `(dE)/(dt)= (mu_(0)e^(2))/(6pic)w^(rarr_2)` For an electron performing (linear) harmonic vibrations `vec(w)` is the some definite direction with `w_(x)=-w^(2)x` say. Thus `(dE)/(dt)=-(mu_(0)e^(2)omega^(4))/(6pic)x^(2)` If the radiation loss is small (i.e., if `omega` is no to large), then the motion of the electron is always close to simple harmonic wiht slowly decreasing amplitude. Then we can write `E=(1)/(2)m omega^(2)a^(2)` and `x=a cos omegat` and avarage the above equation ignoring the variation of `a` in any cycle. Thus we get the equation, on using `lt x^(2) ge(1)/(2)a^(2)` `(dE)/(dt)=-(mu_(0)e^(2)omega^(4))/(6pic).(1)/(2)a^(2)=-(mu_(0)e^(2)omega^(2))/(6pi mc)E` since `E=(1)/(2)momega^(2)a^(2)` for a harmonic oscillator. This equation intergates to `E=E_(o)e^(-t//T)` where `T= 6pi mc//e^(2)omega^(2)mu_(0)` It is than see that energy decreases `eta` times in `t_(0)=T In eta=(6pi mc)/(e^(2)omega^(2)mu_(0))In eta= 14.7ns` |
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| 111. |
Evaluate the amount of heat produced during a day by a `beta^(-)`- active `Na^(24)` preparation of mass `m= 1.0 mg`. The beta particles are assumed to possess an average kinetic energy equal to `1//3` of the highest possible energy of the given decay. The half-life of `Na^(24)` is `T= 15 hours`. |
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Answer» The masses are `Na^(24)= 24-0.00903 amu` and `Mg^(24)= 24-0.01496 amu` The reaction is `Na^(24) rarr Mg^(24)+e^(-)+bar(v)_(e )` The maximum `K.E` of electrons is `0.00593xx93MeV= 5.52MeV` Average `K.E.` according to the problem is then `(5.52)/(3)= 1.84MeV` The initial number of `Na^(24)` is `(10^(-3)xx6.023xx10^(23))/(24)=2.51xx10^(19)` The fraction decying in a day is `1-(2)^(-24//15)= 0.67` Hence the heat produced in a day is `0.67xx2.51xx10^(19)xx1.84xx1.602xx10^(-13)` Joul `= 4.95MJ` |
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| 112. |
The activity of a certain preparation decreases `2.5` times after `7.0` days. Find its half time. |
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Answer» If the half-life is `T` days `(2)^(-7//T)=(1)/(2.5)` Hence `(7)/(T)= (In 2.5)/(In 2)` or `T=(7In 2)/(In 2.5)= 5.30 days` |
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| 113. |
At the initial moment the activity of a certain radionuclie totalled `650` particles per minute. What will be the activity of the preperation after half its half-life period? |
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Answer» The activity is proprotional to the number to the number of parent nuclei(assuming that the daughter is not radioactive). In half its half-life period, the number of parent nucli decrease by a factor `(2)^(-1//2)=(1)/(sqrt(2))` So activity decreases to `(650)/(sqrt(2))= 460` particles per minute. |
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| 114. |
Four lowest energy levels of `H`-atom are shown in the figure. The number of possoble emission lines would be A. 3B. 4C. 5D. 6 |
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Answer» Correct Answer - D |
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| 115. |
Assuming the propagation velocities of longitudinal and transverse vibrations to be the same and equal to `v`, find the Debye temperature (a) for a unidimensional crystal, i.e., a chain identical atoms, incorporating `n_(0)` atoms per unit length, (b) for a two-dimensional crystal, i.e., a plance square grid consisting of identical atoms, containing `n_(0)` atoms per unit area, (c ) for a simple cubic lattice consisting of identical atoms, containing `n_(0)` atoms per unit volume. |
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Answer» To detemine the Debye temperature we cut off the high frequency modes in such a way as to get the total number of modes correctly. (a) In a linear crystal with `n_(0)l` atoms, the number of modes of transverse vibrations in any given plane cannot exceed `n_(0)l`. Then `n_(0)l=(l)/(pi v) int_(0)^(omega_(0)) d omega=(l)/(piV)omega_(0)` The cut off frequency `omega_(0)` is related to the Debye temperature `Theta` by ` ħ omega_(0)=k Theta` Thus `Theta=(( ħ)/(k))pin_(0)v` (b) In a square lattice, the number of modes of transverse oscillations cannot exceed `n_(0)S` Thus `n_(0)S=(S)/(2 piv^(2)) int_(0)^(omega_(0)) omega d omega=(S)/(4pi v^(2))omega_(0)^(2)` or `Theta=(ħ)/(k)omega_(0)=((ħ)/(k))(sqrt(4pin_(0)))v` In a cubic crystal, the maximum number of transverse waves must be `2n_(0)V` (two for each atom). Thus `2n_(0)V=(V)/(pi^(2)v^(3))int_(0)^(omega_(0)) pmega^(2) d omega=(V omega_(0)^(3))/(3 pi^(2)v^(3))` Thus `Theta=((ħ)/(k))v(6pi^(2)n_(0))^(1//3)` |
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| 116. |
Calculate the Debye temperature for iron in which the propagation velocities of longtitudinal and transverse vibrations are equal ot `5.85` and `3.23Km//s` respectivel. |
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Answer» We proceed as in the prevoius example. The total number of modes must be `3n_(0)v` (total trnasverse and one longitudinal per atom). On the other hand of transverse modes per unit frequency interval is by `dN^(_|_)=(Vomega^(2))/(pi^(2)V_(_|_)^(3))d omega` while the number of longitudinal moder per unit frequency interval is given by `dN^(||)=(V omega^(2))/(2pi^(2)v_(||)^(3))d omega` The total number per unit frequency interval is `dN=(V omega^(2))/(2pi^(2))((2)/(V_(_|_)^(3))+(1)/(V_(||)^(3)))d omega` If the high frequency cut off is at `omega_(0)=(kTheta)/( ħ)`, the total number of modes will be `3n_(0)V=(V)/(6 pi^(2))((2)/(V_(_|_)^(3))+(1)/(v_(||)^(3)))((k Theta)/( ħ))^(3)` Here `n_(0)` is the number of iron atoms per unit volume. Thus `: Theta=(ħ)/(k)[18 pi^(2)n_(0)//((2)/(v_(_|_)^(3))+(1)/(v_(||)^(3)))]^(1//3)` For iron `n_(0)=N_(A)//(M)/(rho)=(rhoN_(A))/(M)` `(rho=` density, `M=` atomic weight of iron `N_(A)=` Avogardo number). `n_(0)= 8.389xx10^(22)per c c` Substituting the data we get `Theta= 469.1K` |
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| 117. |
What element has a hygrogen-like specturm whose lines have wavelength four times shorter than those of atomic hydeogen? |
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Answer» If all wavelength are four times shorter but otherwise similar to the hydrogen atom specturm then the energy levels of the given atom must be four times greater. This means `E_(n)= -(4E_(H))/(n^(2))` compared to `E_(n)= -(E_(H))/(n^(2))` for hydrogen atom. Therefore the specturm is that of `He^(+)ion(Z-2)` |
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| 118. |
In a middle of the rotatation-vibration band of emission specturm of `HCI` molecule, where the "Zeroth" line is forbidden by the selcetion rules, the interval between neighbouring lines is `Delta omega= 0.79.10^(13) s^(-1)`. Calculate the distance between the nuclei of an `HCI` molecule. |
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Answer» In the ratation vibration band the main transition is due to change in vibrational quantum number `v rarr v-1`. Together with this roatational quantum number may change. The "Zeroeth line" `0 rarr 0` is forbidden in this case so the neighbouring lines arise due to `1 rarr 0 or 0 rarr 1` in the rotaional quantum number. Now `E=E_(v)+(ħ^(2))/(2I)J(J+1)` Thus `ħ omega=ħ omega_(0)+(ħ^(2))/(2I)(+-2)` Hence `Delta omega =(2ħ)/(I)=(2V)/(mud^(2))` so `d= sqrt((2ħ)/(muDelta omega))` Substitution gives `d= 0.128nm`. |
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| 119. |
In a chain of identical atoms the vibration frequency `omega` depends on wave number `k`as `omega= omega_(max) sin (ka//2)`, where `omega_(max)` is the maximum vibration frequency `omega`, a is the distance between neighbouring atoms. Making use of this dispersion relation, find the dependence of the number of longitudinal vibrations per unit. frequency interval on `omega` i.e., `dN//omega`, if the length of the chain is `l`. Having obtained `dN//omega`, find the total number `N` of possible longitudinal vibrations of the chain. |
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Answer» If the chain has `N` atoms, we can assume atom number `0` and `N+1` held ficed. Then the displacement of the `n^(th)` atom has the form `u_(n)=A("sin"(m pi)/(L),n a) sin omega t` Here `k=(m pi)/(L)`. Allowed frequencies then have the form `omega=omega_(max) "sin"(ka)/(2)` In our from only `+vek` values are allowed. The number of models in a wave number range `dk` is `dN=(Ldk)/(pi)=(L)/(pi)(dK)/(d omega)` But `d omega=(a)/(2) omega_(max)"cos"(ka)/(2)dk` Hence `(d omegaqa)/(dk)=(a)/(2)sqrt(omega_(max)^(2)-omega^(2))` So `dN=(2L)/(pia)(d omega)/(sqrt(omega_(max)^(2)-omega^(2)))` (b) The total number modes is `N= int_(0)^(omega_(max))(2L)/(pi a)(d omega)/(sqrt(omega^(2)_(max)-omega^(2)))=(2L)/(pi a).(pi)/(2)=(L)/(a)` i.e., the number of atoms in the chain. |
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| 120. |
The first line of the sharp series of atomic cesium is a doublet with wavelengths `1358.8` and `1469.5 nm`. Find the frquency intervals (in rads/s units) between the components of the sequent lines of that series. |
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Answer» The sharp series arise from the transitions `ns rarr mp`. The `s` lines are unsplit so the splitting is due entirely to the `p` level. The frequency difference between sequen t lines is `(DeltaE)/( ħ)` and is the same for all lines of the sharp series. It is `(1)/( ħ)((2pi ħc)/(lambda_(1))-(2pi ħc)/(lambda_(2)))=(2pi cDeltalambda)/(lambda_(1)lambda_(2))` Evaluation gives `1.645xx10^(14)rad//s` |
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| 121. |
Find the natural vibration frequency and the quasielatic force coefficient of an `S_(2)` molecule if the wavelength of the red and voilet satellites, closests to the fixed line, in the vibration specturm of Raman scattering are equal to `346.6` and `330.0 nm`. |
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Answer» As in the previous problem `omega= pic((1)/(lambda_(V))-(1)/(lambda_(R)))=(pi c(lambda_(R)-lambda_(V)))/(lambda_(R)lambda_(V))= 1.368xx10^(14)rad//s` The force constant `x` is defined by `x= mu omega^(2)` where `mu=` reduced mass of the `S_(2)` molecule. Substitution gives `x=5.01N//cm` |
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| 122. |
Find the number of natural transverse vibration of a rigtht-angled parallelipoped of volume `V` in the frquency interval from `omega to omega +d omega` if the propogation velocity of vibrations is equal to `v` |
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Answer» For transverse vibrations of a 3-diamensional continuum (in the from of a cube say). We have the equation `(del vec(xi))/(delt^(2))=V^(2)vecgrad^(2)vec(xi), di v vec(xi)=0` Here `vec(xi)=vec(xi)(x,y,z,t)`. We look for solutions in the form `vec(xi)=vec(A) sink_(1)x,sink_(2)y,sink_(3Ƶ), sin (omega t+delta)` This requires `omega^(2)=v^(2)(k_(1)^(2)+k_(2)^(2)+k_(3)^(2))` From the boundary condition that `vec(xi)=0` for `x=0, x=l,y=0,y=l,Ƶ=l`, we get `k_(1)=(n_(1)pi)/(l),k_(2)=(n_(2)pi)/(l)=(n_(2)pi)/(l),k_(3)=(n_(3)pi)/(l)` where `n_(1):n_(2),n_(3)` are nonzero positive intergers. We then get `n_(1)^(2)+n_(2)^(2)+n_(3)^(2)=((lomega)/(pi v))^(2)` Each triplet `(n_(1),n_(2),n_(3))` determines a possible mode and the number of such modes whose frequency `le omega` is the volume of the all positive octant of a sphere of radius `(l omega)/(pi v)`. Considering also the fact that subsidiary conditions div `vec(xi)=0` implies two independent values of `vec(A)` for we find `N(omega)=(1)/(8).(4 pi)/(3)((l omega)/(pi v))^(3).2=(V omega^(3))/(3 pi^(2)v^(3))` Thus `dN=(Vomega^(2))/(pi^(2)v^(3))domega` |
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| 123. |
If the following atoms and molecylates for the transition from `n = 2` to `n = 1`, the spectral line of minimum wavelength will be produced byA. Hydrogen atomB. Deuterium atomC. Uni-ionized heliumD. di-ionized lithium |
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Answer» Correct Answer - D |
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| 124. |
Calculate the mean lifetime of excited atoms if it is known that the intensity of the spectral line appering due to transition to the ground state diminishes by a factor `eta=25` over a distance `l=2.5mm` along the stream of atoms whose velocity is `v=600m//s`. |
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Answer» Let `T=` mean life time of the excited atoms. Then the number of exicted atoms will decrease with time as `e^(-t//T)`. In time `t` the atom travels a distance `vt` so `t=(l)/(v)`. Thus the number of excited atoms a in a beam that has travesed a distance `l` has decreased by `e^(-l//vT)` The intensity of the line is proportional to the number of excited atoms in the beam. Thus `e^(-l//vtau)=(1)/(eta) or tau=(l)/(V i n eta)=1.29xx10^(-16)` second. |
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| 125. |
The electron in a hydrogen atom makes a transition from an excited state to the ground state. Which of the following statements is true?A. Its kinetic energy increases and its potential and total energies decreaseB. Its kinetic energy decreases, potential energy increases and its total energy remains the sameC. Its kinetic and total energies decrease and its potential energy increasesD. Its kinetic, potential and total energies decreases |
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Answer» Correct Answer - A |
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| 126. |
For an `HCI` molecule find the rotational quantum numbers of two neighbouring levels whose energies differ by `7.86 meV`. The nuclei of the molecule are separated by distacne of `127.5p m`. |
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Answer» The axis of rotation passes through the centre of mass of the `HCl` molecule.The distance of the two atoms from the centre of mass are `d_(H)=(m_(cl))/(M_(Hcl))d, =(m_(H))/(m_(Hcl))d` Thus `I=` moment of interita about the axis `(4)/(2) m_(H)d_(H)^(2)+m_(cl)d_(cl)^(2)=(m_(H)M_(cl))/(m_(H)+m_(cl))d^(2)` The energy difference between two neighbouring levels whose quantum numbers are `J&J-1` is `( ħ^(2))/(2I).2J=( Jħ^(2))/(I)= 7.86 meV` Hence `J=3` and the levels have quantum numbers `2 & 3`. |
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| 127. |
Find the number of ratational levels per unit energy interval, `dN//dE`, for a diatomic molecule as a function of rotational energy `E`. Calculate of an magnitude for an iodine molecule in the state with rotational quantum number `J=10`. The distance between the nuclei of that molecule is equal to `267 p m`. |
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Answer» From the formula `J(J+1)( ħ^(2))/(2I)= E` we get `J(J+1)- 2IE// ħ^(2)` or `(J+(1)/(2))^(2)-(1)/(4)= (2IE)/( ħ^(2))` Hence `J=-(1)/(2)+sqrt((1)/(4)+(2IE)/( ħ^(2)))` writing `J+1= -(1)/(2)+sqrt((1)/(4)+(2I)/( ħ^(2))(E+DeltaE))` we find `1=sqrt((1)/(4)+(2I)/(ħ^(2))E+(2I)/(ħ^(2))DeltaE)-sqrt((1)/(4)+(2IE)/(ħ^(2)))` `sqrt((1)/(4)+(2I)/(ħ^(2))E)[(1+(DeltaE)/(E+(ħ^(2))/(8I)))^(1//2)]` `sqrt((1)/(4)+(2I)/(ħ^(2))E).(DeltaE)/(2(E+(ħ^(2))/(8I)))` `sqrt((2I)/(ħ^(2)))(DeltaE)/(2sqrt((E+(ħ^(2))/(8I))))` The quantity `(dN)/(dE)is (1)/(DeltaE)`. For large `E` it is `(dN)/(dE)= sqrt((I)/(2 ħ^(2)E))` For an iodine molecule `I= M_(I)d^(2)//2= 7.57xx10^(-38)gm cm^(2)` Thus for `J= 10` `(dN)/(dE)=sqrt((I)/(2ħ^(2).(ħ^(2))/(2I)J(J+1)))=(I)/(sqrt(J(J+1)ħ^(2)))` Substitution gives `(dN)/(dE)= 1.04xx10^(4)` levels per `eV`. |
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| 128. |
A hydrogen atom (ionisation potential `13.6 eV`) makes a transition from third excited state to first excied state. The enegry of the photon emitted in the process isA. 1.89 eVB. 2.55 eVC. 12.09 eVD. 12.75 eV |
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Answer» Correct Answer - B |
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| 129. |
Find the ratio of energies required to excite a diatomic molecule to the first vibrational and to the first rotational level. Calculate that ratio for the following molecules: Molecule, `omega, 10^(14s^(-1)` d, pm (a) `H_(2) 8.3 74` (b) `HI 4.35 160` (c ) `I_(2) 040 267` Here `omega` is the natural vibration frequency of a molecule, `d` is the distance between nuclei. |
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Answer» For the first rotational level `E_(rot)= 2(ħ^(2))/(2I)=(ħ^(2))/(I)` and for the first vibrational level `E_(vib)= .ħ omega` Thus `xi=(E_(vib))/(E_(rof))=(Iomega)/(ħ)` Here `omega=` frequency of vibration. Now `I= mud^(2)=(m_(1)m_(2))/(m_(1)+m_(2))d^(2)` (a) For `H_(2)` moleculeI `=4.58xx10^(-41)gm cm^(2)` and `xi= 36` (b) For HI moleculeI `=4.2xx47xx10^(-40) gm cm^(2)` and `xi= 175` (c ) For `I_(2)` moleculeI `=7.57xx10^(-38) gm cm^(2)` and `xi= 2872` |
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| 130. |
The minimum enegry required to excite a hydrogen atom from its ground state isA. 13.6 eVB. `- 13.6 eV`C. 3.4 eVD. 10.2 eV |
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Answer» Correct Answer - D |
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| 131. |
Ratio of the wavelength of first line of Lyaman series and first line of Balmer series isA. `1 : 3`B. `27 : 5`C. `5 : 27`D. `4 : 9` |
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Answer» Correct Answer - C |
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| 132. |
Find the Rydberg correction for the `3P` term of a `Na` atom whose first exitation potential is `2.10V` and whose valance electron in the normal `3S` state has the binding energy `5.14 eV`. |
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Answer» The enrgy of the `3p` state must be `-(E_(0)-e varphi)` where `-E_(0)` is the energy of the `3S` state. Then `E_(0)-evarphi_(1)=( ħR)/((3+alpha_(1))^(2))` so `alpha_(1)=sqrt(( ħR)/(E_(0)-e varphi_(1)))-3 =-0.885` |
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| 133. |
The binding energy of the valance electron in a `Li` atom in the states `2S` and `2P` is equal to `5.39` and `3.54 eV` respectively. Find the Rydberg corrections for `s` and `p` terms of the atom. |
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Answer» From the Rydberg formula we write `E_(n)= -( ħR)/((n+alpha_(l)^(2))` we use ` ħR= 13.6eV`. Then for n= 2 state `5.39= -(13.6)/((2+alpha_(0))^(2)), l=0(S)` state `alpha_(0)~~ -0.41` for `p` state `3.54= -(13.6)/((2+alpha_(1))^(2))` `alpha_(1)=-0.039`. |
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| 134. |
An atom possesing the total angular momentum ` ħvv6` is in the state with spin quantum number `S=1`. In the corresponding vector model the angle between the spin momentum and the total angular momentum is `theta= 73.2^(@)`.Write the spectral symbol for the term that state. |
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Answer» Total angular momentum ` ħsqrt(6)` means `J=2`. It is gives that `S=1` This means that `L=1,2 or 3`. From vector model relation `L(L+1) ħ^(2)=6 ħ^(2)+2 ħ^(2)sqrt(6)sqrt(2) cos 73.2^(@)` `=5.998 ħ^(2)~~6 ħ^(2)` Thus `L=2` and the spectral symbol of the state is `.^(3)D_(2)` |
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| 135. |
An atom is in the state whose multiplities is three and the total angular momentum is `ħsqrt(20)`. What can the corresponding quantum number `L` be equal to? |
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Answer» Multiplicity is `2S+1 so S=1` Total angular momentum is ` ħsqrt(J(J+1))` so `J=4`. Then `L` must be equal `3,4,5` in order that `J=4` may be included in `|L-S| to L + S` |
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| 136. |
A valance electron is known in a sodium atom is in the state with principle quantum number`n=3`, with the total angular momentum being the greatest possible, What is its magnetic moment in that state? |
| Answer» A valence electron in a sodium atom is in the state with principal quantum number `n=3`, with the total angular momentum being the greatest possible, What is its magnetic moment in that state? | |
| 137. |
Find the total angular momentum of an atom in the state with `S=3//2` and `L=2` if its magnetic moment is known to be equal to zero. |
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Answer» Since `mu=0` we must have either `J=0` or `g=0` is incomplete with `L=2` and `S=(3)/(2)`. Hence `g=0`. Thus `0=1+(J(J+1)+(3)/(2)xx(5)/(2)-2xx3)/(2J(J+1))` or `-3J(J+1)=(15)/(4)-6=-(9)/(4)` Hence `J=(1)/(2)` Thus `M= ħsqrt((1)/(2)xx(3)/(2))=( ħsqrt(3))/(2)` |
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| 138. |
In the Bohr model of a hydrogen atom, the centripetal force is furnished by the Coulomb attraction between the proton and the electrons. If `a_(0)` is the radius of the ground state orbit, `m` is the mass and `e` is the charge on the electron and `e_(0)` is the vacuum permittivity, the speed of the electron is |
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Answer» Correct Answer - C |
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| 139. |
A certain atom is in the state in which `S=2`, the total angular momentum `sqrt(2 ħ)`, and the magnetic moment is equal to zero. Write the spectral symbol of the corresponding term. |
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Answer» From `M= ħsqrt(J+1)=sqrt(2 ħ)` we find `J=1`. From the zero value of the magnetic moment we find `g=0` or `1+(1xx2L(L+1)+2xx3)/(2xx1xx2)=0` `1+(-L(L+1)+8)/(4)=0` or `12=L(L+1)` Hence `L=3`. The state is `.^(5)F_(1)`. |
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| 140. |
Which of the following spectral series in hydrogen atom give spectral line of `4860 Å`A. ) LymanB. BalmerC. PaschenD. Brackett |
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Answer» Correct Answer - B |
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| 141. |
Amongst `alpha,beta` and `gamma-`particles, `alpha-`particle has maximum penetrating power. The `alpha-`particle is heavier than `beta` and `gamma-`particle.A. If both assertion and reason are true and the reason is the correct explanation of the assertionB. If both assertion and reason are true but reason is not the correct explanation of the assertionC. If assertion is true but reason is false.D. If the assertion and reason both are false. |
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Answer» Correct Answer - D |
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| 142. |
The ionising power of `alpha-`particle is less compared to `alpha-`particles but their penetrating power is more. The mass of `beta-`particle is less than the mass of `alpha`-particle.A. If both assertion and reason are true and the reason is the correct explanation of the assertionB. If both assertion and reason are true but reason is not the correct explanation of the assertionC. If assertion is true but reason is false.D. If the assertion and reason both are false. |
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Answer» Correct Answer - B |
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| 143. |
As par Bohr model, the minimum energy (in `eV`) required to remove an electron from the ground state of doubly ionized `Li` atom `(Z = 3)` isA. 1.51B. 13.6C. 40.8D. 122.4 |
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Answer» Correct Answer - D |
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| 144. |
A relativistic particle with rest mass `m` cillides with a stationary particle of mass `M` and activities at reaction leading to formation of new particles: `m+Mrarrm_(1)+m_(2)+…..`, where the rest masses of newly formed particles are written on the right-hand side. Making use of invariance of the quantity `E^(2)-p^(2)c^(2)`, demonstrate that the threshold kinetic energy of the particle `m` required for this reaction is defined by Eq.(6.7c). |
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Answer» With particle masses standing for the names of particles, the reaction is `m+Mrarrm_(1)+m_(2)+…..` On R.H.S. let the energy momenta be `(E_(1),cvec(p)_(1)),(E_(2),cvec(p)_(2))` etc. On the left the energy momentum of the particle `m is (E,cvec(p))` and that of the particle is `(Mc^(2),vec(O))`, where ofcourse, the usual relations `E^(2)-c^(2)vec(p)^(2)=m^(2)c^(4)eta` hold. From the conservation of energy momentum we see that `(E+Mc^(2))^(2)-c^(2)vec(p)^(2)=(SigmaE_(i))^(2)-(SigmaE_(i))^(2)-(Sigmacvec(p)i)^(2)` Left hand side of `m^(2)c^(4)+M^(2)c^(4)+2Mc^(2)E` We evaluate the R.H.S. in the frame where `Sigmavec(pi)=0` (`CM` frame of the decay product). Then R.H.S `=(SigmaE_(i))^(2)=(SigmaE_(i))^(2) ge(Sigmam_(i)c^(2))^(2)` beacuse all energies are `+ve`. Therefore we have the result `E ge((Sigmam_(i))^(2)-m^(2)-M^(2))/(2M)c^(2)` M or Since `E=mc^(2)+T`, we see that `T geT_(th)` where `T_(th)=((Sigmam_(i))^(2)-(m+M)^(2))/(2M)c^(2)` |
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| 145. |
Find the threshold energy of gamma quantum required to form (a) an electron-positron pair in the field of a stationary electron, (b) a pair of pions of opposite signs in the field of a stationary proton. |
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Answer» The fromula of problem 3.02 gives `E_(th)=((Sigmam_(i))^(2)-M^(2))/(2M)c^(2)` when the projectile is a photon (a) For `gamma+e^(-)rarre^(-)+e^(-)+e^(+)` `E_(th)=(8m_(e )^(2)-m_(e )^(2))/(2m_(e ))c^(2)= 4m_(e )c^(2)= 2.04MeV` (b) For `gamma+prarrp+pi^(+)pi^(-)` `E_(th)=((M_(p)+2m_(pi))^(2)-M_(P)^(2))/(2M_(p))=(4m_(pi)M_(p)+4m_(pi)^(2))/(2M_(P))c^(2)=2(m_(x)+(m_(x)^(2))/(M_(P)))c^(2)= 320.8MeV` |
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| 146. |
The nuclide `.^131 I` is radioactive, with a half-life of `8.04` days. At noon on January `1`, the activity of a certain sample is `60089`. The activity at noon on January `24` will beA. 75 BqB. Less than 75 BqC. More than 75 BqD. 150 Bq |
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Answer» Correct Answer - C |
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| 147. |
Radioactive element decays to form a stable nuclide, then the rate of decay of reactant `((d N)/(d t))` will vary with time `(t)` as shown in figure.A. B. C. D. |
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Answer» Correct Answer - C |
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| 148. |
Radioactivity of `108` undecayed radioactive nuclei of half life of `50` days is equal to that of `1.2 xx 108` number of undecayed nuclei of some material with half life of `60` days Radioactivity is proportional to half-life.A. If both assertion and reason are true and the reason is the correct explanation of the assertionB. If both assertion and reason are true but reason is not the correct explanation of the assertionC. If assertion is true but reason is false.D. If the assertion and reason both are false. |
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Answer» Correct Answer - C |
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| 149. |
Number of nuclei of a radioactive substance are `1000` and `900` at times `t=0` and time `t=2 s`. Then, number of nuclei at time `t=4s` will beA. 800B. 810C. 790D. 700 |
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Answer» Correct Answer - B |
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| 150. |
A hydrogen atom emits a photon corresponding to an electron transition from `n = 5` to `n = 1`. The recoil speed of hydrogen atom is almost (mass of proton `~~1.6 xx 10^(-27) kg)`.A. 10 msB. `2 xx 10 ms`C. `4 ms`D. `8 xx 10 ms` |
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Answer» Correct Answer - C |
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